Split string into INT array in c [closed] - c

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I got string containing numbers separated by spaces. Numbers can be single-digit, two-digit, or perhaps more-digit. Check the example.
"* SEARCH 2 4 5 12 34 123 207"
I don't know how long the string is (how many numbers it contains), so I cant initiate the array properly. The result should look like this:
array = {2,4,5,12,34,123,207}
Do you have any ideas how to perform this?

Answer 1

like this:
#include <stdio.h>
#include <stdlib.h>
int main(void){
char *input = "* SEARCH 2 4 5 12 34 123 207";
int len = 0;
sscanf(input, "%*[^0-9]%n", &len);//count not-digits(The Number isn't negative)
char *p = input + len;
char *start = p;
int v, n = 0;
while(1 == sscanf(p, "%d%n", &v, &len)){
++n;//count elements
p += len;
}
int array[n];//or allocate by malloc(and free)
char *endp = NULL;
int i;
for(i = 0; i < n; ++i){
array[i] = strtol(start, &endp, 10);
start = endp + 1;
}
//check print
for(i = 0; i < n; ++i)
printf("%d ", array[i]);
puts("");
return 0;
}

Answer 2

You can try this approach. It uses a temporary buffer to hold the current integer that is being processed. It also uses dynamic arrays, to deal with different lengths of the string you want to process, and expands them when necessary. Although using strtok Would be better in this situation.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int
main(int argc, char *argv[]) {
char message[] = "* SEARCH 2 4 5 12 34 123 207";
char *buffer = NULL;
int *integers = NULL;
int buff_size = 1, buff_len = 0;
int int_size = 1, int_len = 0;
int ch, messlen, i, first_int = 0;
/* creating space for dynamic arrays */
buffer = malloc((buff_size+1) * sizeof(*buffer));
integers = malloc(int_size * sizeof(*integers));
/* Checking if mallocs were successful */
if (buffer == NULL || integers == NULL) {
fprintf(stderr, "Malloc problem, please check\n");
exit(EXIT_FAILURE);
}
messlen = strlen(message);
/* going over each character in string */
for (ch = 0; ch < messlen; ch++) {
/* checking for first digit that is read */
if (isdigit(message[ch])) {
first_int = 1;
/* found, but is there space available? */
if (buff_size == buff_len) {
buff_size++;
buffer = realloc(buffer, (2*buff_size) * sizeof(*buffer));
}
buffer[buff_len++] = message[ch];
buffer[buff_len] = '\0';
}
/* checking for first space after first integer read */
if (isspace(message[ch]) && first_int == 1) {
if (int_size == int_len) {
int_size++;
integers = realloc(integers, (2*int_size) * sizeof(*integers));
}
integers[int_len] = atoi(buffer);
int_len++;
/* reset for next integer */
buff_size = 1;
buff_len = 0;
first_int = 0;
}
/* for last integer found */
if (isdigit(message[ch]) && ch == messlen-1) {
integers[int_len] = atoi(buffer);
int_len++;
}
}
printf("Your string: %s\n", message);
printf("\nYour integer array:\n");
for (i = 0; i < int_len; i++) {
printf("%d ", integers[i]);
}
/* Being careful and always free at the end */
/* Always a good idea */
free(integers);
free(buffer);
return 0;
}

Answer 3

You can read each character and verify if it is in range of >=48(Ascii of 0) and less than = 57(Ascii of 9). If so is the case read them into a array Otherwise you could copy them to a temporary string and convert to int using functions like atoi()
#include <stdio.h>
int main(int argc, char *argv[])
{
int j=0,k,res;
char buff[10];
while(str[j])
{
if((str[j]>='0')&&(str[j]<='9'))
{
k=0;
while((str[j]!=' ')&&(str[j]!='\0'))
{
buff[k]=str[j++];
k++;
}
buff[k]=0;
res=atoi(buff);
//Store this result to an array
}
j++;
}
return 0;
}